Explanation Given: The probability density function is f ( x ) = 1 28 ( 2 x + 3 ) over [ 0 , 4 ] . Concept used: The probability density function f be defined on the interval [ a , b ] associated with a continuous random variable X . Then the expected value μ of X is written as follows. μ = ∫ a b x f ( x ) d x ...... (1) The variance of X is written as follows. Var ( X ) = ∫ a b ( x − μ ) 2 f ( x ) d x ...... (2) The standard deviation of X is written as follows. σ = Var ( X ) ...... (3) Calculation: The probability density function is given as f ( x ) = 1 28 ( 2 x + 3 ) over [ 0 , 4 ] . Substitute 0 for a , 4 for b and 1 28 ( 2 x + 3 ) for f ( x ) in equation ( 1 ) to obtain the mean value as shown below. μ = ∫ 0 4 x ( 1 28 ( 2 x + 3 ) ) d x = 1 28 ∫ 0 4 x ( 2 x + 3 ) d x = 1 28 ∫ 0 4 ( 2 x 2 + 3 x ) d x = 1 28 [ 2 x 3 3 + 3 x 2 2 ] 0 4 Further simplify the above equation as follows. μ = 1 28 [ 2 ( 4 ) 3 3 + 3 ( 4 ) 2 2 − ( 2 ( 0 ) 3 3 + 3 ( 0 ) 2 2 ) ] = 1 28 [ 66

Applied Calculus for the Managerial, Life, and Social Sciences

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ISBN: 9781305887831

Author: Tan

Publisher: CENGAGE CO

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Chapter 10, Problem 14RE

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The mean, variance and standard deviation of the random variable x associated with the probability density function f(x)=128(2x+3) over [0,4].

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Chapter 10, Problem 13RE

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Chapter 10 Solutions

Applied Calculus for the Managerial, Life, and Social Sciences

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The mean, variance and standard deviation of the random variable x associated with the probability density function f ( x ) = 1 28 ( 2 x + 3 ) over [ 0 , 4 ] .

Solution Summary: The author explains the probability density function f be defined on the interval left[a,bright] associated with a continuous random variable

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The mean, variance and standard deviation of the random variable x associated with the probability density function f(x)=128(2x+3) over [0,4].

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Chapter 10 Solutions